A boy uses a rope to haul a box of weight 600N up an inclined wooden plane of effective length 3.0m and onto a platform 1.0m high. If in doing so, he exerts a force of 400N on the rope. Draw the diagram of all the forces acting on the box and calculate the frictional force between the box and the incline plane
sinA = 1m/3m = 0.333. A = 19.5o.
Fp = Mg*sinA = 600*sin19.5 = 200 N. = Force parallel with plane.
Fs = Force of static or starting friction.
Fe-Fp-Fs = M*a.
400-200-Fs = M*0 = 0,
Fs = 200 N.
correct
To draw the diagram of all the forces acting on the box, we must consider the forces of gravity, normal force, and the force applied by the boy.
1. Gravity: The force of gravity acts vertically downwards, with a magnitude of 600N. We can represent this force as a downward arrow labeled as "600N".
|
↓ 600N
2. Normal Force: The normal force acts perpendicular to the inclined plane. Since the box is on an incline, the normal force will be inclined as well, but its magnitude will be equal and opposite to the force of gravity. So, the normal force can be represented as an upward arrow, labeled as "600N".
↑ 600N
|
3. Force Applied by the Boy: The boy exerts a force of 400N on the rope in order to haul the box up the incline. The force applied by the boy is directed parallel to the incline, and we can represent it as an upward arrow labeled as "400N".
| 400N
↑
Now, to calculate the frictional force between the box and the inclined plane, we need to determine the force components parallel and perpendicular to the plane.
4. Force Component Parallel to the Plane: Since the force applied by the boy is parallel to the incline, this force component will have the same magnitude as the force applied by the boy, i.e., 400N.
5. Force Component Perpendicular to the Plane: This force component is the difference between the weight of the box (600N) and the force applied by the boy (400N). Therefore, the force component perpendicular to the plane is 600N - 400N = 200N, acting perpendicular to the incline.
Now, we can calculate the frictional force using the force component parallel to the plane.
Frictional force = force component parallel to the plane = 400N
Therefore, the frictional force between the box and the incline plane is 400N.
To draw the diagram of the forces acting on the box, we need to consider the different forces involved. Here's a breakdown:
1. Weight of the box (600N): This force acts vertically downwards, towards the center of the Earth.
2. Normal force (N): This force acts perpendicular to the surface of the incline, pushing upwards. It balances the weight of the box and can be calculated using the formula N = mg, where m is the mass of the box and g is the acceleration due to gravity. In this case, the normal force is equal to the weight of the box since it is not accelerating vertically.
3. Applied force (400N): This force is exerted by the boy on the rope. It acts at an angle on the incline and is responsible for moving the box up the incline.
4. Frictional force (f): This force opposes the motion of the box and acts parallel to the incline in the opposite direction to the applied force. It can be calculated using the formula f = μN, where μ is the coefficient of friction between the box and the incline.
Now, let's calculate the frictional force between the box and the incline:
Since the incline is at an angle, we need to resolve the forces into their horizontal and vertical components. The component of the weight perpendicular to the incline is equal to N, while the component parallel to the incline is mg sinθ, where θ is the angle of the incline.
Using trigonometry, the angle θ can be found as sinθ = opposite/hypotenuse = 1.0m/3.0m = 1/3. Therefore, sinθ = 1/3, and θ = arcsin(1/3) = 19.47 degrees.
The component of the weight parallel to the incline is mg sinθ = 600N * sin(19.47) ≈ 204.34N.
Since the box is moving up the incline, the applied force (400N) must overcome the frictional force. Therefore, we can say that the frictional force is equal to the applied force: f = 400N.
So, the frictional force between the box and the incline is approximately 400N.